Question

help pleaseeeeee -
find the derivative:
f(x)=arcsinx/arccosx

Answer 1

would it just be:\[\frac{ 1 }{ \sqrt{1-x^{2}} } \over \frac{ 1 }{ \sqrt{1-x^{2}} }\]

Answer 2

denominator should be a negative

Answer 3

This can be rewriten as:\[\sin (f(x)\arccos x)=x\]Take the derivative on both sides and you get:\[1=\cos(f(x)\arccos x)(f'(x)\arccos x+f(x) (\arccos x)')\]\[f'(x)=\frac{1}{\cos(f(x)\arccos x)\arccos x}-\frac{f(x)(\arccos x)'}{\arccos x}\]\[f'(x)=\frac{1}{\cos(\arcsin x)\arccos x}-\frac{ \arcsin x (\arccos x)'}{\arccos^2 x}\]Now we need to fin the derivative of arccos x, we do the same thing to find it.
\[g(x)=\arccos x\]\[\cos(g(x))=x\]Taking the derivative:\[-\sin(g(x))g'(x)=1\]\[g'(x)=-\frac{1}{\sin(\arccos x)}\]Now putting it on the formula for f' we get:\[f'(x)=\frac{1}{\cos(\arcsin x)\arccos x}+\frac{\arcsin x}{\sin(\arccos x)\arccos^2 x}\]

Answer 4

why cant i use my inverse trig formulas?

Answer 5

????

Answer 6

Well, you can but I don't like remembering formulas because we can forget them, I prefer remembering only the necessary and nothing else, but if you know them you can use for sure.

Answer 7

Use quotient rule with your formulas and you will get the correct answer.

Answer 8

i have another question if you dont mind.

Answer 9

Oh, I only saw your other post now, you need to apply the chain rule or the quotient rule to do this, its not just taking the derivative of each and divide them.

Answer 10

Sure, ask it.

Answer 11

okay... f(x)=arcsin(3x/square root of 9x^2+4

Answer 12

i drew my triangle to look like thjis

Answer 13

cause it said to draw a triangle to make it easier.
but i dont know where to go from here.

Answer 14

What is it asking?

Answer 15

find the derivative of the function

Answer 16

it said to draw a right triangle to get a simplier expression instead of doing the derivative directly.

Answer 17

Knowing the 2 sides of the triangle you can get the third, and using arccos instead of arcsin simplifies it in this case.

Answer 18

so cos^-1 = 2/3x .. now what do i do?

Answer 19

Actually, I meant arctan, because 3x/2 is the tangent, then you just need to use the formula for the tangent and multiply the result by 3/2